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The next set of medium/hard DS number properties questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2) $$2\sqrt{x^2}$$ is a prime number.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205341

2. If a positive integer n has exactly two positive factors what is the value of n?

(1) n/2 is one of the factors of n
(2) The lowest common multiple of n and n + 10 is an even number.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205355

3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

(1) Both x and y is have 3 positive factors.
(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers

Solution: https://gmatclub.com/forum/new-set-numb ... l#p1209015

4. Each digit of the three-digit integer K is a positive multiple of 4, what is the value of K?

(1) The units digit of K is the least common multiple of the tens and hundreds digit of K
(2) K is NOT a multiple of 3.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205361

5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number.
(2) c! is a prime number

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205364

6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative
(2) The product of the smallest and largest integers in the list is a prime number.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205373

7. Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205378

8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number
(2) The product of any two terms of the set is a terminating decimal

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205382

9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

(1) ab = 2
(2) 0 < a < b < 2

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205389

10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

(1) 9 is NOT a factor of N
(2) 125 is a factor of N

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205392

BONUS QUESTION:
11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205398

Kudos points for each correct solution!!!
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6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

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2] n has only two positive factorss, therefore n is prime.

(1) n/2 is one of the factors of n. If n is prime, and n/2 is a factor of n, then n = 2. Sufficient

(2) As Bunuel always says, the greatest rule about LCM is that for two numbers a,b, we have: a.b =LCM(a,b).GCF(a,b). If n is prime, let's suppose n>2 (in this case, n will have to be odd since it is prime). If n is odd, then n+10 is odd as well. From the rule we just mentioned:

n.(n+10) = odd*odd = odd = LCM.GCF -> Therefore, LCM(n,n+10) cannot be even (otherwise, the product would be even).

So, n has to be 2. Sufficient

##### General Discussion
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11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then $$2-y\leq{0}$$ and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

So, we have that: $$|x - 2| < 1$$, which implies that $$-1 < x-2 < 1$$, or $$1 < x < 3$$, thus $$x=2=prime$$. Sufficient.

(2) x + y - 3 = |1-y|. Since y is a positive integer, then $$1-y\leq{0}$$, thus $$|1-y|=-(1-y)$$. So, we have that $$x + y - 3 = -(1-y)$$, which gives $$x=2=prime$$. Sufficient.

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7. Is x the square of an integer?

The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).

Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

(1) When x is divided by 12 the remainder is 6. Given that $$x=12q+6=6(2q+1)=2*3*(2q+1)$$. Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.

(2) When x is divided by 14 the remainder is 2. Given that $$x=14p+2$$. So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.

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8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Theory:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
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3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

Notice that since x and y are consecutive perfect squares, then $$\sqrt{x}$$ and $$\sqrt{y}$$ are consecutive integers.

(1) Both x and y have 3 positive factors. This statement implies that $$x=(prime_1)^2$$ and $$y=(prime_2)^2$$. From above we have that $$\sqrt{x}=prime_1$$ and $$\sqrt{y}=prime_2$$ are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, $$x=(prime_1)^2=4$$ and $$y=(prime_2)^2=9$$. The remainder when 9 is divided by 4 is 1. Sufficient.

(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers. The same here: $$\sqrt{x}=2$$ and $$\sqrt{y}=3$$. Sufficient.

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2. If a positive integer n has exactly two positive factors what is the value of n?

Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself.

(1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient.

(2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient.

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5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that:
A. The factorial of a negative number is undefined.
B. 0!=1.
C. Only two factorials are odd: 0!=1 and 1!=1.
D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

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SOLUTIONS:

1. If x is an integer, what is the value of x?

(1) $$|23x|$$ is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) $$2\sqrt{x^2}$$ is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

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9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, =2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that $$0\leq{a}<1$$ and $$1\leq{b}<2$$ (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

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4. Each digit of the three-digit integer K is a positive multiple of 4, what is the value of K?

(1) The units digit of K is the least common multiple of the tens and hundreds digit of K. K could be 444, 488, 848, or 888. Not sufficient.

(2) K is NOT a multiple of 3. K could be 448, 484, 488, 844, 848, or 884. Not sufficient.

(1)+(2) From above K could be 488 or 848. Not sufficient.

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3]

(1) Both x and y have 3 factors. Which means that x and y are not only perfect squares, but also perfect squares of PRIME numbers. Let's remember that the only 2 consecutive prime numbers are 2 and 3, therefore: Sufficient

(2) This already gives us directly the same information that sqrt(x)=2, and sqrt(y) = 3, therefore: Sufficient

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10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

$$N = 3^x*5^y$$ has 12 positive factors means that (x+1)(y+1)=12=2*6=6*2=3*4=4*3. We can have the following cases:

$$N = 3^1*5^5$$;
$$N = 3^5*5^1$$;
$$N = 3^2*5^3$$;
$$N = 3^3*5^2$$.

(1) 9 is NOT a factor of N. This implies that the power of 3 is less than 2, thus N could only be $$3^1*5^5$$. Sufficient.

(2) 125 is a factor of N. This implies that the power of 5 is more than or equal to 3, thus N could be $$3^1*5^5$$ or $$3^2*5^3$$. Not sufficient.

THEORY.
Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
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Re: New Set: Number Properties!!!  [#permalink]

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imhimanshu wrote:
Bunuel wrote:
11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then $$y-2\leq{0}$$ and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

Hi Bunuel,
Thanks for the awesome questions that you ve posted.

I have a doubt in the above question. Could you please elaborate why can't we solve the Statement A as below -

|x - 2| < 2 - y ---> y-2< X-2 < 2-y

I tried solving above, by removing the Modulus, however I am not getting any specific value.

Kindly respond.

Thanks

Responding to a pm:

If you must solve it using algebra, then do this:

|x - 2| < 2 - y

Take the two cases:

Case 1:
x - 2 >= 0 i.e. x >= 2
x - 2 < 2 - y
x + y < 4
Since x must be 2 or greater than 2 and x and y both are positive integers, x must be 2 and y must be 1 (x cannot be 3 or greater since y must be at least 1 but the sum of x and y in that case will not be less than 4).

Case 2:
x - 2 < 0 i.e. x < 2
The only value that x can take in this case is 1.
|1 - 2| < 2 - y
1 < 2 - y
y < 1
But we know that y must be a positive integer so x cannot take value 1.

The only value that x can take is 2 which is prime. Hence this statement alone is sufficient.

Mind you, using algebra too you will have to use logic so you might as well look at Bunuel's solution again.
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Re: New Set: Number Properties!!!  [#permalink]

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11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|

We know that x >0 and y>0 and they are integers.

From F.S 1, we have 2-y>=0 or y<=2. Thus y can only be 2 or 1. Now if y=2, we would have 0>some thing positive or 0>0(when x also equal to 2). Either case is not possible. Thus, y can only be 1. For y=1, we can only have x = 2. Which is prime. Sufficient.

From F.S 2, we have either y>1 or y<1. Now as y is a positive integer, y can't be less than 1.For y=1, we anyways have x=2(prime).In the first case, we have y>1--> x+y-3 = y-1 or x=2(prime). Thus, Sufficient.

D.
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Re: New Set: Number Properties!!!  [#permalink]

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7.Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2

From F.S 1, we have x = 12q+6 --> 6(2q+1). For x to be a square of an integer, we should have 2q+1 of the form 6^pk^2, where both q,p and k are integers and p is odd. Now we know that 2q+1 is an odd number and 6^pk^2 is even. Thus they can never be equal and hence x can never be the square of an integer. Sufficient.

From F.S 2, we have x = 14q+2 --> 2(7q+1).Just as above, we should have 7q+1 = 2^pk^2. Now for q=1, k=2 and p=1, we have 8=8, thus x is the square of an integer. But for q=0, x is not. Insufficient.

Basically for the second fact statement, we can plug in easily. No need for the elaborate theory.

A.
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5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number
(2) c! is a prime number

From F.S 1, we have b! = odd, thus b can be 0 or 1.But, as factorial notation is only for positive integers, thus, if b=0, then a would become negative and thus b is only equal to 1.Now, a can only be 0 as we are given that a! exists. But nothing has been mentioned about c. All we know is that c>1 and an integer. Insufficient.

From F.S 2, we have c! is a prime number. Again, c has to be positive and c can only be 2.However, a and b can take any values, even negative. Insufficient.

Taking both together, we have a=0, b=1 and c=2. Sufficient.

C.
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Re: New Set: Number Properties!!!  [#permalink]

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2. If a positive integer n has exactly two positive factors what is the value of n?

(1) n/2 is one of the factors of n
(2) The lowest common multiple of n and n + 10 is an even number.

n is a positive integer that has exactly two positive factors --> 1 must be one of its factor --> n is a prime number and not equal to 1 (because 1 has only one positive factors, itself)

So these two factors could be (1,2) or (1,3) or (1,5) ... and n could be 2,3,5 .......

(1) n/2 is one of the factors of n

n/2 is a factor of n --> n/2 is an integer and from the pairs (1,2), (1,3) ... only 2 is divisible by 2
Hence , n = 2 --> (1) SUFFICIENT

(2) The lowest common multiple of n and n + 10 is an even number.

LCM (n,n+10) = EVEN

If n = 2 then LCM(2,12) = 12, which is EVEN
If n = 3 then LCM (3,13) = 39, which is ODD
Except for n=2, Like n=3, n = 5,7,11 .... LCM (n,n+10) will be ALWAYS ODD.
Hence, n = 2 --> (2) SUFFICIENT

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Re: New Set: Number Properties!!!  [#permalink]

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3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

(1) Both x and y is have 3 positive factors.
(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers

(1) Both x and y is have 3 positive factors.

Consecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 ....
Among these numbers, only 4 and 9 are consecutive perfect squares that have 3 positive factors ( for instance 16 = 4*4 = 2*2*2*2 --> 5 factors and SO ON )
Hence, y=9 and x=4 --> 9 = 4.2 +1 --> R = 1

Thus, (1) SUFFICIENT

(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers
Consecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 ....
Among these numbers, only 4 and 9 are consecutive perfect squares that have their square roots as prime numbers ( for example : $$\sqrt{16}$$ = 4, which is not a prime number and SO ON )
Hence, y=9 and x=4 --> 9 = 4.2 --> R = 1

Thus, (2) SUFFICIENT

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Originally posted by Rock750 on 26 Mar 2013, 03:57.
Last edited by Rock750 on 29 Mar 2013, 04:20, edited 1 time in total. Re: New Set: Number Properties!!!   [#permalink] 26 Mar 2013, 03:57

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